Signal Amplifier with 2N2222

Posted on Oct 8, 2012

A small signal amplifier, perhaps of the type to follow the previous buffer amplifier. We will assume we are buffering and amplifying our signal from the voltage controlled oscillator tutorial. In those examples we were generating and buffering 1.8 to 2.0 Mhz signals for the 160M band. The configuration is much the same as other class `A` amplifier designs covered in previous tutorials. The output circuit consists of a low pass filter network which also converts the desired output impedance we want Q1 to see to our standard 50 ohms output. The 100 ohm resistor, RFC XL2 and the 0.01 uF capacitors are purely for decoupling purposes i.e., to keep RF out of the small signal amplifier power supply as well as other stages. Let's consider firstly the input circuit of our small signal amplifier.

Signal Amplifier with 2N2222
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Q1 is biased for DC conditions by R1, R2 and the emitter resistor of 270 ohms in this instance. Alert readers will be aware I like to bias the base voltage of my transistors to about 25% of Vcc (.25 * 12V) or 3V. It follows then that R1 will be about 3 times the value of R2 - think about it!. If the base voltage is around 3V then the emitter voltage is going to be 3v - 0.65V = 2.35V. Don't follow that? Go back to class "A" amplifier designs covered in previous tutorials. If the emitter voltage is 2.35V approx. then the emitter current Ie through the emitter resistor of 270 ohms must be (from ohms law) 2.35 / 270 = 0.0087 or 8.7 mA. I've also said elsewhere I like base current to be about 1/7th of emitter current - alright these are my foibles and others would disagree. They're welcome to write their own papers. So base current is going to be about 1 mA and seeing R1 + R2 are connected across 12V it follows that (from ohms law) R1 + R2 = 12V / .001 = 12,000 ohms or 12K. For biasing R1 is 3 times R2 so using simple maths R2 is 25% or 3K and R1 would be 9K which are not necessarily readily available standard values. We will make R2 = 3K3 and R1 = 10K which if you do all your sums is near enough and probably about a third of the values others might use. So we have our DC conditions satisfied and the 0.01 capacitor in parallel with the emitter resistor means for RF purposes the emitter is at ground potential. This then...

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